For what value of k does the quadratic equation
(k +4) x2 + (k +1) x +1 = 0 have equal roots? (CBSE 2013)
Let D be the discriminant of the given equation.
We have a = k + 4, b = k + 1, c = 1
D = b2 – 4ac
= (k +1)2 – 4(K + 4)(k + 1)
= k2 + 1 + 2k – 4(k2 + k + 4k + 4)
= = k2 + 1 + 2k – 4k2 - 4k - 16k – 16
= k2 – 2k – 15
= k2 – 5K + 3K – 15
= k(k – 5) + 3(k – 5)
= (k – 5) ( k + 3)
If roots are real, then
D = 0
⇒ (k – 5)(k + 3) = 0
⇒ k = 5,-3
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